Question

Given the parametric equations, x = 2t + 5 and y = 3t^2 - 1, find the point on the graph at time, t = 4.

Answer 1

Answer: (13, 47)

The expression t = 4 means that 4 units of time have come off the clock (possibly 4 seconds). Plug this value into each equation given for x and y

x = 2*t + 5 = 2*4 + 5 = 8 + 5 = 13

y = 3*t^2 - 1 = 3*4^2 - 1 = 3*16 - 1 = 48 - 1 = 47

In short, if t = 4 then it leads to x = 13 and y = 47 at the same time. These x and y values pair up to get the final answer (13,47)